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∫cos(2 cot-1​​{ root(1-x/1+x)} = ????

  • ∫cos(2 cot-1​​{ root(1-x/1+x)} = ????

Grade:12

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
Hello Student,
Please find answer to your question below

I = \int cos(2cot^{-1}\sqrt{\frac{1-x}{1+x}})dx
x = cos\theta
dx = -sin\theta d\theta
I = \int cos(2cot^{-1}\sqrt{\frac{1-cos\theta }{1+cos\theta }})(-sin\theta )d\theta
I = \int cos(2cot^{-1}(tan\frac{\theta }{2}))(-sin\theta )d\theta
I = \int cos(2cot^{-1}(cot(\frac{\pi }{2}-\frac{\theta }{2}))(-sin\theta )d\theta
I = \int cos(\pi -\theta )(-sin\theta )d\theta
I = \int -cos(\theta )(-sin\theta )d\theta
I = \frac{1}{2}\int 2cos(\theta )(sin\theta )d\theta
I = \frac{1}{2}\int (sin2\theta )d\theta
I = \frac{-cos2\theta }{4} + c
I = \frac{-cos(2cos^{-1}x) }{4} + c
Ritika Das
105 Points
4 years ago
Hello I am a student. 
I = -cos(2cos-1x)/4  +  c  should have been further expressed as follows:-
I = -(2cos2theta-1)/4  +  c
or, I = -(2x2-1)/4  +  c
or, I = -(x2)/2  +  c

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